question 16\nopen - box problem. an open - box (top open) is made from a rectangular material of dimensions…

question 16\nopen - box problem. an open - box (top open) is made from a rectangular material of dimensions a = 13 inches by b = 12 inches by cutting a square of side x at each corner and turning up the sides (see the figure). determine the value of x that results in a box the maximum volume.\n(1) express the volume v as a function of x (do not simplify): v=\n(2) determine the upper limit of the domain of the function: v\n(3) simplify the function: v=\n(4) the maximum volume is v=\n(5) answer the original question. the value of x that maximizes the volume is:\nwatch the video starting at 5:35\nquestion help: video post to forum\nsubmit question

question 16\nopen - box problem. an open - box (top open) is made from a rectangular material of dimensions a = 13 inches by b = 12 inches by cutting a square of side x at each corner and turning up the sides (see the figure). determine the value of x that results in a box the maximum volume.\n(1) express the volume v as a function of x (do not simplify): v=\n(2) determine the upper limit of the domain of the function: v\n(3) simplify the function: v=\n(4) the maximum volume is v=\n(5) answer the original question. the value of x that maximizes the volume is:\nwatch the video starting at 5:35\nquestion help: video post to forum\nsubmit question

Answer

Explanation:

Step1: Find volume function

When a square of side - length $x$ is cut from each corner of the rectangle with length $a = 13$ inches and width $b = 12$ inches and the sides are turned up, the length of the box is $l=13 - 2x$, the width is $w = 12-2x$, and the height is $h=x$. So the volume function $V(x)=(13 - 2x)(12 - 2x)x$.

Step2: Determine domain upper - limit

The length and width of the box must be non - negative. So $13−2x\geq0$ and $12 - 2x\geq0$. Also, $x\geq0$. Solving $12−2x = 0$ gives $x = 6$. Solving $13−2x=0$ gives $x=\frac{13}{2}=6.5$. The smaller of these non - negative roots for the non - negativity of length and width gives the upper limit of the domain. So the upper limit of the domain of $V(x)$ is $x = 6$.

Step3: Simplify the function

[ \begin{align*} V(x)&=(13 - 2x)(12 - 2x)x\ &=(156-26x-24x + 4x^{2})x\ &=(156-50x + 4x^{2})x\ &=156x-50x^{2}+4x^{3} \end{align*} ]

Step4: Find maximum volume

First, find the derivative of $V(x)$: $V^\prime(x)=156-100x + 12x^{2}$. Set $V^\prime(x)=0$, so $12x^{2}-100x + 156 = 0$. Divide through by $4$ to get $3x^{2}-25x + 39 = 0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$, here $a = 3$, $b=-25$, $c = 39$. So $x=\frac{25\pm\sqrt{(-25)^{2}-4\times3\times39}}{2\times3}=\frac{25\pm\sqrt{625 - 468}}{6}=\frac{25\pm\sqrt{157}}{6}$. $x_1=\frac{25+\sqrt{157}}{6}\approx\frac{25 + 12.53}{6}=\frac{37.53}{6}\approx6.26$ (rejected as it is out of the domain $0\leq x\leq6$), $x_2=\frac{25-\sqrt{157}}{6}\approx\frac{25 - 12.53}{6}=\frac{12.47}{6}\approx2.08$. Then find $V(2.08)=156\times2.08-50\times(2.08)^{2}+4\times(2.08)^{3}\approx324.48-50\times4.3264 + 4\times8.9989\approx324.48-216.32+35.9956\approx144.16$.

Step5: Find $x$ for maximum volume

The value of $x$ that maximizes the volume is $x=\frac{25-\sqrt{157}}{6}\approx2.08$.

Answer:

(1) $(13 - 2x)(12 - 2x)x$ (2) $6$ (3) $4x^{3}-50x^{2}+156x$ (4) $\approx144.16$ (5) $\frac{25 - \sqrt{157}}{6}\approx2.08$